The ultimate aim is to solve the following integral: \begin{equation} \label{eq:Icos1} \begin{aligned} I = \int\limits_{0}^{2\pi} \frac{\mathrm{exp}\left(c \cos(\varphi)\right)\mathrm{d}\varphi}{a - b \cos(\varphi)}, \end{aligned} \end{equation} where $a>b$ and $c>0$.

Now with substitutions $z = \mathrm{exp}(i\varphi)$, $\mathrm{d}\varphi = \frac{\mathrm{d}z}{zi}$, $\cos(\varphi) = \frac{1}{2}\left(z+\frac{1}{z}\right)$ we can rewrite the original problem in terms of a complex variable $z$ as \begin{equation} \label{eq:Icos3} \begin{aligned} I = \oint\limits_{C} \frac{\mathrm{exp}\left(\frac{c}{2}\left(z+\frac{1}{z}\right)\right)\mathrm{d}z}{(z-z_1)(z-z_2)}, \end{aligned} \end{equation} where $z_{1,2}$ are simple poles. Clearly, there is a singularity at $z_0 = 0$, however I can't find corresponding residue value.

I've found the solution to an easier problem (below) using complex integration and calculation of residues, however finding the singularity at $z_0 = 0$ is where I stuck. Namely I don't think its possible to express $Res(f,z_0) = a_{-1}$ as $\frac{f(z_1,z_2,a,b,c)}{z}$, as people suggest here in other related questions.

Does anyone have any suggestions on how to find the $Res(f,z_0) = a_{-1}$ where the numerator in the integrand goes to infinity?

Easier problem that I've solved is \begin{equation} \label{eq:Icos2} \begin{aligned} I_{easier} = \int\limits_{0}^{2\pi} \frac{\mathrm{d}\varphi}{a - b \cos(\varphi)} = \frac{2\pi b}{\sqrt{a^2-b^2}}, \end{aligned} \end{equation} for $a>b$.

  • $\begingroup$ I don't know if it rigurous but try L'Hopital, which will get you rid of the $a$ in the denominator and then express the exponential as a series. $\endgroup$ – Jorge Lavín Nov 9 '12 at 13:12

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